2015 IMO Problems/Problem 4
Problem
Triangle has circumcircle
and circumcenter
. A circle
with center
intersects the segment
at points
and
, such that
,
,
, and
are all different and lie on line
in this order. Let
and
be the points of intersection of
and
, such that
,
,
,
, and
lie on
in this order. Let
be the second point of intersection of the circumcircle of triangle
and the segment
. Let
be the second point of intersection of the circumcircle of triangle
and the segment
.
Suppose that the lines and
are different and intersect at the point
. Prove that
lies on the line
.
Proposed by Silouanos Brazitikos and Evangelos Psychas, Greece
Solution
Lemma (On three chords). If two lines pass through different endpoints of two circles' common chord, then the other two chords cut by these lines on the circles are parallel.
Proof The second and the third chords are anti-parallel to the first (common) chord with respect to the given lines, so they are parallel to each other.
To solve this problem, it is sufficient to apply the lemma 5 times. Indeed, let the lines meet
second time at
respectively. One of the circles that figure in lemma is always
, while the other is one of three other circles from the problem statement. Applying the lemma to the lines
and
,
and
,
and
,
and
,
and
, we get
,
,
,
,
, respectively. From this,
,
,
. Therefore,
. This means that
and
are symmetric wrt
, a diameter of
through
. So are
and
, as
. Therefore, the lines
and
are symmetric wrt
and meet on it.
See Also
2015 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |